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Last edited by hbghlyj 2025-3-21 04:13若 $x_{n+1} \ln x_{n+1}=x_n-1, x_1=2$ ，
求证：（1）$x_n>x_{n+1}>1$
（2）$x_1+x_2+\ldots+x_n \geq n+2-2^{1-n}$
首先易证$x_n>1$，这里就不多说了
\[x_n-x_{n+1}-1=x_{n+1}(\ln(x_{n+1})-1)>(\ln(x_{n+1})+1)(\ln(x_{n+1})-1)=\ln^2(x_{n+1})-1\]
\[x_n-x_{n+1}>\ln^2(x_{n+1})>0\]
第二问，硬来就行了，强行证明$x_n\ge 1+2^{1-n}$
考察函数$f(x)=x\ln(x)-2(x-1)$，易证$1\le x\le 2$时$f(x)<0$
因此有
\[x_{n+1}\ln(x_{n+1})=x_n-1<2(x_{n+1}-1)\]
\[x_n-1>2^{1-n}\]
\[x_n>1+2^{1-n}\]
\[\sum_{k=1}^nx_k>\sum_{k=1}^n(1+2^{1-n})=n+2-2^{1-n}\] |
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